Revision history [back]

Here is how you would do it, assuming A and B are NumPy arrays:

import matplotlib.pyplot as plt
mean = A.mean()
standard_deviation = A.std()
plt.hist(B)
plt.savefig('histogram.png')

Unfortunately Sage does not yet have good histogram functionality; however, it does have mean and standard deviation, with the benefit that you can get exact output:

sage: std(range(10))
sqrt(55/6)

Here is how you would do it, assuming A and B are NumPy arrays:

import matplotlib.pyplot as plt
mean = A.mean()
standard_deviation = A.std()
plt.hist(B)
plt.savefig('histogram.png')

Unfortunately Sage does not yet have good histogram functionality; however, it does have mean and standard deviation, with the benefit that you can get exact output:

sage: std(range(10))
sqrt(55/6)

def mean2(data, freq):
    return sum(map(prod, zip(data, freq))) / sum(freq)

def std2(data, freq):
    data2 = []
    for index, item in enumerate(data):
        data2.extend([item] * freq[index])
    return std(data2)

Here is how you would do it, assuming A and B are NumPy arrays:

import matplotlib.pyplot as plt
mean = A.mean()
standard_deviation = A.std()
plt.hist(B)
plt.savefig('histogram.png')

Unfortunately Sage does not yet have good histogram functionality; however, it does have mean and standard deviation, with the benefit that you can get exact output:

sage: std(range(10))
sqrt(55/6)

EDIT: Here's a way to take the (integer) frequencies into account:

def mean2(data, freq):
    return sum(map(prod, zip(data, freq))) / sum(freq)

def std2(data, freq):
    data2 = []
    for index, item in enumerate(data):
        data2.extend([item] * freq[index])
    return std(data2)